Respuesta :
x^3−x^2−12x=0
x(x^2-x-12)=0
x(x^2-4x+3x-12)=0
x(x(x-4)+3(x-4))=0
x(x+3)(x-4)=0
So the "zeros" are:
x=(-3,0,4)
x(x^2-x-12)=0
x(x^2-4x+3x-12)=0
x(x(x-4)+3(x-4))=0
x(x+3)(x-4)=0
So the "zeros" are:
x=(-3,0,4)